Four simple harmonic vibrations:
$y_1 = 8 \cos(\omega t)$;
$y_2 = 4 \cos(\omega t + \frac{\pi}{2})$;
$y_3 = 2 \cos(\omega t + \pi)$;
$y_4 = 1 \cos(\omega t + \frac{3\pi}{2})$,
are superposed on each other. The resulting amplitude and phase are respectively:

$A$ particle executes two types of $SHM$. $x_1 = A_1 \sin \omega t$ and $x_2 = A_2 \sin [\omega t + \frac{\pi}{3}]$.
$(a)$ Find the displacement at time $t = 0$.
$(b)$ Find the maximum speed of the particle.
$(c)$ Find the maximum acceleration of the particle.

Two simple harmonic motions are represented by the equations $x_{1}=5 \sin \left(2 \pi t+\frac{\pi}{4}\right)$ and $x_{2}=5 \sqrt{2}(\sin 2 \pi t+\cos 2 \pi t)$. The ratio of the amplitude of $x_{1}$ and $x_{2}$ is

Two simple harmonic motions are represented by $y_1 = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$ and $y_2 = 5 \sin [2 \pi t + \frac{\pi}{4}]$. The ratio of their amplitudes is

Two particles execute simple harmonic motion along the same straight line with the same amplitude and same frequency. The two particles pass one another when moving in opposite directions each time at a distance of $\frac{1}{\sqrt{2}}$ times the amplitude from their common mean position. The phase difference between the two particles is (in $^{\circ}$)

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to $A$ and $T$,respectively. At time $t=0$,one particle has displacement $A$ while the other one has displacement $\frac{-A}{2}$ and they are moving towards each other. If they cross each other at time $t$,then $t$ is